Wednesday, 26 October 2016

real analysis - Showing that fracsqrt[n]n!n rightarrowfrac1e




Show:limnnn!n=1e



So I can expand the numerator by geometric mean. Letting Cn=(ln(a1)+...+ln(an))/n. Let the numerator be called an and the denominator be bn Is there a way to use this statement so that I could force the original sequence into the form of 1/(1+1n)n


Answer



I would like to use the following lemma:




If limnan=a and an>0 for all n, then we have
limnna1a2an=a




Let an=(1+1n)n, then an>0 for all n and limnan=e. Applying () we have
e=limnna1a2an=limnn(21)1(32)2(n+1n)n=limnn(n+1)nn!=limnn+1nn!=limnnnn!+limn1nn!=limnnnn!


where we use (1) in the last equality to show that
limn1nn!=0.



It follows from (2) that
limnnn!n=1e.


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